Open Query 10

course Phys 202

I just took the MCAT this past weekend and studying for it kind of took over my life. I know I'm behind, but I will catch up by the August 10th deadline. I just wanted to let you know why my work stopped for awhile.

010. `query 9 *********************************************

Question: Query introductory set 6, problems 1-10

explain how we know that the velocity of a periodic wave is equal to the product of its wavelength and frequency

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Your Solution:

We know that v = f*lamda because f is the peaks per second that pass a given point, while lamda is the distance between two peaks in meters. This makes f*lamda pass a certain point every second giving us the velocity in meters/second.

f = peaks/second

lamda = m/ peaks

f* lamda = peaks/sec * m/peak = m/sec = velocity

Confidence rating: 3

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Given Solution:

** we know how many wavelength segments will pass every second, and we know the length of each, so that multiplying the two gives us the velocity with which they must be passing **

Your Self-Critique: OK

Your Self-Critique rating: OK

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Question: explain how we can reason out that the period of a periodic wave is equal to its wavelength divided by its velocity

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Your Solution:

We know from the precious example that f = v/lamda

Period is just 1/f so it is lamda (wavelength)/v

or m/m/sec = sec^ -1 = period

Confidence rating: 3

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Given Solution:

** If we know how far it is between peaks (wavelength) and how fast the wavetrain is passing (velocity) we can divide the distance between peaks by the velocity to see how much time passes between peaks at a given point. That is, period is wavelength / velocity. **

Your Self-Critique: OK

Your Self-Critique rating: OK

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Question: explain why the equation of motion at a position x along a sinusoidal wave is A sin( `omega t - x / v) if the equation of motion at the x = 0 position is A sin(`omega t)

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Your Solution:

You have to include a factor to take into account the lag time in which it takes to get from x position 0 to the new position. This accounts for the – x/v in the parenthesis. This is the time lab because you are subtracting the position x divided by its velocity which gives you the excess time. m divided by m/s = seconds

Confidence rating: 3

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Given Solution:

** the key is the time delay. Time for the disturbance to get from x = 0 to position x is x / v. What happens at the new position is delayed by time x/v, so what happens there at clock time t happened at x=0 when clock time was t = x/v.

In more detail: If x is the distance down the wave then x / v is the time it takes the wave to travel that distance. What happens at time t at position x is what happened at time t - x/v at position x=0.

That expression should be y = sin(`omega * (t - x / v)).

The sine function goes from -1 to 0 to 1 to 0 to -1 to 0 to 1 to 0 ..., one cycle after another. In harmonic waves the motion of a point on the wave (think of the motion of a black mark on a white rope with vertical pulses traveling down the rope) will go thru this sort of motion (down, middle, up, middle, down, etc.) as repeated pulses pass.

If I'm creating the pulses at my end, and that black mark is some distance x down in rope, then what you see at the black mark is what I did at time x/v earlier. **

STUDENT COMMENT (University Physics):

According to the Y&F book (p.553) we get the expression for a sinusoidal wave moving the the +x-direction with the equation:

Y(x,t) = A*cos[omega*(t-x/v)]

I am not sure where the sine came from in the equation in the question. The book uses the cosine function to represent the waves motion.

The choice of the cosine function is arbitrary. Either function, or a combination of both, can come out of the solution to the wave equation (that's the partial differential equation which relates the second derivative with respect to position to the second derivative with respect to time).

The sine and cosine functions differ only by a phase difference of 90 degrees, and either can be used to describe simple harmonic motion or the motion of harmonic waves. The choice simply depends on the initial conditions of the system.

We don't want to get into solving the wave equation here, but the point can be illustrated by considering simple harmonic motion, which is characterized by F_net = - k x (leading to m x '' = - k x or x '' = -k/m * x, where derivatives are with respect to time).

The general solution to the equation x '' = - k / m * x is x = B sin(omega t) + C cos(omega t), where B and C are arbitrary constants and omega = sqrt(k/m).

B sin(omega t) + C cos(omega t) = A sin(omega t + phi), where A and phi are determined by B, C and the choice to use the sine function on the right-hand side.

B sin(omega t) + C cos(omega t) = A cos(omega t + phi), where A and phi are determined by B, C and the choice to use the cosine function on the right-hand side. The value of A will be the same as if we had used the sine function on the right, and the value of phi will differ by 90 degrees or pi/2 radians.

Your Self-Critique: OK

Your Self-Critique rating: OK

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Question: Query introductory set six, problems 11-14

given the length of a string how do we determine the wavelengths of the first few harmonics?

Answered in Query for 11

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Question: Given the wavelengths of the first few harmonics and the velocity of a wave disturbance in the string, how do we determine the frequencies of the first few harmonics?

Answered in Query 11

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Question: Given the tension and mass density of a string how do we determine the velocity of the wave in the string?

Answered in Query 11

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Question: gen phy explain in your own words the meaning of the principal of superposition

Answered in Query 11

Question: gen phy what does it mean to say that the angle of reflection is equal to the angle of incidence? Answered in Query 11

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&#Very good work. Let me know if you have questions. &#

I understand about the MCAT. If you think of it, let me know how you did overall (and on the physics part, for which I hope this course has been helpful; I also suspect you've been through a review course and learned a significant amount of physics there).